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最新推荐文章于 2019-04-26 19:59:55 发布

翻译最新推荐文章于 2019-04-26 19:59:55 发布 · 310 阅读

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#output #go

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16974 Accepted Submission(s): 7374

Problem Description

A simple mathematical formula for e is

where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output

  
   n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

Source

Greater New York 2000

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#include<stdio.h>

long factorial(int n){
    long result=1;
    for(int i=n;i>0;i--)
        result*=i;
    return result;
}

int main(void){
    printf("n e\n- -----------\n");
    double sum=0;
    for(int i=0;i<10;i++){
        sum+=(double)1/factorial(i);
        if(i<2)
            printf("%d %.0lf\n",i,sum);
        else if(i==2)
            printf("%d %.1lf\n",i,sum);
        else 
            printf("%d %.9lf\n",i,sum);
    }
    return 0;
}